9.3 THE SIMPLEX METHOD: MAXIMIZATION
For linear programming problems involving two variables, the graphical solution method introduced in Section 9.2 is convenient. However, for problems involving more than two variables or problems involving a large number of constraints, it is better to use solution methods that are adaptable to computers. One such method is called the simplex method, developed by George Dantzig in 1946. It provides us with a systematic way of examining the vertices of the feasible region to determine the optimal value of the objective function.
We introduce this method with an example.
Suppose we want to find the maximum value of where and subject to the following constraints.
Since the left-hand side of each inequality is less than or equal to the right-hand side, there must exist nonnegative numbers and that can be added to the left side of each equa- tion to produce the following system of linear equations.
The numbers and are called slack variables because they take up the “slack” in each inequality.
s3 s1,s2
2x11 5x21 s11 s2 1 s35 90 2x115x21 s11 s21 s35 27 2x115x21 s11 s21 s35 11
s3 s1,s2 2x11 5x2 # 90
2x115x2 # 27 2x115x2 # 11
x2$ 0, x1 $ 0
z5 4x11 6x2,
Standard Form of a Linear Programming Problem
A linear programming problem is in standard form if it seeks to maximize the objec-
tive function subject to the constraints
.. .
where and After adding slack variables, the corresponding system of constraint equations is
.. .
where si $ 0.
am1x11 am2x2 1 . . . 1 amnxn 1 sm5 bm
a21x11 a22x21 . . . 1 a2nxn 1 s2 5 b2
a11x11 a12x21 . . . 1 a1nxn1 s1 5 b1
bi $ 0.
xi$ 0
am1x11 am2x2 1 . . . 1 amnxn # bm
a21x11 a22x21 . . . 1 a2nxn # b2
a11x11 a12x21 . . . 1 a1nxn #b1 z5 c1x11 c2x21 . . . 1 cnxn
R E M A R K: Note that for a linear programming problem in standard form, the objective function is to be maximized, not minimized. (Minimization problems will be discussed in Sections 9.4 and 9.5.)
A basic solution of a linear programming problem in standard form is a solution of the constraint equations in which at most m variables are nonzero––the variables that are nonzero are called basic variables. A basic solution for which all variables are nonnegative is called a basic feasible solution.
The Simplex Tableau
The simplex method is carried out by performing elementary row operations on a matrix that we call the simplex tableau. This tableau consists of the augmented matrix corre- sponding to the constraint equations together with the coefficients of the objective function written in the form
In the tableau, it is customary to omit the coefficient of z. For instance, the simplex tableau for the linear programming problem
Objective function
is as follows.
Basic
x1 x2 s1 s2 s3 b Variables
1 1 0 0 11 s1
1 1 0 1 0 27 s2
2 5 0 0 1 90 s3
0 0 0 0
↑
Current z–value
For this initial simplex tableau, the basic variables are and and the nonbasic variables (which have a value of zero) are and . Hence, from the two columns that are farthest to the right, we see that the current solution is
and
This solution is a basic feasible solution and is often written as sx1,x2, s1, s2, s3d 5 s0, 0, 11, 27, 90d.
s35 90.
x15 0, x25 0, s15 11, s25 27, x2 x1
s3, s1,s2, 26
24 21
2x11 5x21 s11 s21 s35 90 2x115x21 s11 s21 s35 27 2x115x21 s11 s21 s35 11 z5 4x11 6x2
2c1x1 2 c2x2 2 . . . 2 cnxn1 s0ds11 s0ds21 . . . 1 s0d sm1 z 5 0.
sx1, x2, . . . , xn, s1, s2, . . . , smd
}
ConstraintsThe entry in the lower–right corner of the simplex tableau is the current value of z. Note that the bottom–row entries under and are the negatives of the coefficients of and
in the objective function
To perform an optimality check for a solution represented by a simplex tableau, we look at the entries in the bottom row of the tableau. If any of these entries are negative (as above), then the current solution is not optimal.
Pivoting
Once we have set up the initial simplex tableau for a linear programming problem, the sim- plex method consists of checking for optimality and then, if the current solution is not op- timal, improving the current solution. (An improved solution is one that has a larger z-value than the current solution.) To improve the current solution, we bring a new basic variable into the solution––we call this variable the entering variable. This implies that one of the current basic variables must leave, otherwise we would have too many variables for a basic solution––we call this variable the departing variable. We choose the entering and departing variables as follows.
1. The entering variable corresponds to the smallest (the most negative) entry in the bottom row of the tableau.
2. The departing variable corresponds to the smallest nonnegative ratio of , in the column determined by the entering variable.
3. The entry in the simplex tableau in the entering variable’s column and the departing variable’s row is called the pivot.
Finally, to form the improved solution, we apply Gauss-Jordan elimination to the column that contains the pivot, as illustrated in the following example. (This process is called pivoting.)
E X A M P L E 1 Pivoting to Find an Improved Solution
Use the simplex method to find an improved solution for the linear programming problem represented by the following tableau.
Basic
x1 x2 s1 s2 s3 b Variables
1 1 0 0 11 s1
1 1 0 1 0 27 s2
2 5 0 0 1 90 s3
0 0 0 0
The objective function for this problem is z5 4x11 6x2. 26
24 21
biyaij
z5 4x11 6x2. x2
x1 x2
x1
Solution Note that the current solution corresponds to a z–value of 0. To improve this solution, we determine that is the entering variable, because is the smallest entry in the bottom row.
Basic
x1 x2 s1 s2 s3 b Variables
1 1 0 0 11 s1
1 1 0 1 0 27 s2
2 5 0 0 1 90 s3
0 0 0 0
↑
Entering
To see why we choose as the entering variable, remember that Hence, it appears that a unit change in produces a change of 6 in z, whereas a unit change in produces a change of only 4 in z.
To find the departing variable, we locate the ’s that have corresponding positive elements in the entering variables column and form the following ratios.
Here the smallest positive ratio is 11, so we choose as the departing variable.
Basic
x1 x2 s1 s2 s3 b Variables
1 1 0 0 11 s1 ←Departing
1 1 0 1 0 27 s2
2 5 0 0 1 90 s3
0 0 0 0
↑
Entering
Note that the pivot is the entry in the first row and second column. Now, we use Gauss- Jordan elimination to obtain the following improved solution.
Before Pivoting After Pivoting
The new tableau now appears as follows.
3
2102127 1000 212516 0100 0010 111635664 3
212412 26115 1000 0100 0010 11279004
26 24 21
s1 11
1 5 11, 27
1 5 27, 90 5 5 18
bi
x1 x2
z5 4x11 6x2. x2
26 24 21 26
x2
sx15 0, x25 0, s15 11, s25 27, s35 90d
Basic
x1 x2 s1 s2 s3 b Variables
1 1 0 0 11 x2
2 0 1 0 16 s2
7 0 0 1 35 s3
0 6 0 0 66
Note that has replaced in the basis column and the improved solution
has a z-value of
In Example 1 the improved solution is not yet optimal since the bottom row still has a negative entry. Thus, we can apply another iteration of the simplex method to further im- prove our solution as follows. We choose as the entering variable. Moreover, the small-
est nonnegative ratio of and is 5, so is the departing
variable. Gauss-Jordan elimination produces the following.
Thus, the new simplex tableau is as follows.
Basic
x1 x2 s1 s2 s3 b Variables
0 1 0 16 x2
0 0 1 6 s2
1 0 0 5 x1
0 0 0 116
In this tableau, there is still a negative entry in the bottom row. Thus, we choose as the entering variable and s2as the departing variable, as shown in the following tableau.
s1
10
287 7
1
257 7
227 3
7
1 7 2
7
3
0010 1000 2227375787 0100 2107172717 11616654 3
2102121 1000 2121657 0100 00017 11166654 3
2102127 1000 212516 0100 0010 111635664
s3 35y7 5 5
11ys21d, 16y2 5 8, x1 z5 4x11 6x25 4s0d 1 6s11d 5 66.
sx1,x2, s1, s2, s3d 5 s0, 11, 0, 16, 35d s1
x2 210
25 21 21
Basic
x1 x2 s1 s2 s3 b Variables
0 1 0 16 x2
0 0 1 6 s2 ← Departing
1 0 0 5 x1
0 0 0 116
↑ Entering
By performing one more iteration of the simplex method, we obtain the following tableau.
(Try checking this.)
Basic
x1 x2 s1 s2 s3 b Variables
0 1 0 12 x2
0 0 1 14 s1
1 0 0 15 x1
0 0 0 132 ← Maximum z-value
In this tableau, there are no negative elements in the bottom row. We have therefore deter- mined the optimal solution to be
with
R E M A R K: Ties may occur in choosing entering and/or departing variables. Should this happen, any choice among the tied variables may be made.
Because the linear programming problem in Example 1 involved only two decision vari- ables, we could have used a graphical solution technique, as we did in Example 2, Section 9.2. Notice in Figure 9.18 that each iteration in the simplex method corresponds to moving from a given vertex to an adjacent vertex with an improved z-value.
The Simplex Method
We summarize the steps involved in the simplex method as follows.
z5 132 z5 116
z5 66 z5 0
s15, 12d s5, 16d
s0, 11d s0, 0d
z5 4x11 6x25 4s15d 1 6s12d 5 132.
sx1, x2, s1, s2, s3d 5 s15, 12, 14, 0, 0d
2 3 8 3
213 5 3
223 7 3
1
223 3 10
287 7
1
257 7
227 3
7
1 7 2
7
Figure 9.18
x1 x2
5 5 10 15 20 25
10 15 20 25 30 (0, 0)
(0, 11) (5, 16)
(15, 12)
(27, 0)
Note that the basic feasible solution of an initial simplex tableau is
This solution is basic because at most m variables are nonzero (namely the slack variables).
It is feasible because each variable is nonnegative.
In the next two examples, we illustrate the use of the simplex method to solve a problem involving three decision variables.
E X A M P L E 2 The Simplex Method with Three Decision Variables Use the simplex method to find the maximum value of
Objective function
subject to the constraints
where and
Solution Using the basic feasible solution
the initial simplex tableau for this problem is as follows. (Try checking these computations, and note the “tie” that occurs when choosing the first entering variable.)
sx1,x2, x3, s1, s2, s3d 5 s0, 0, 0, 10, 20, 5d x3 $ 0.
x1 $ 0, x2 $ 0, 2x11 2x21 2x3 # 25 2x11 2x22 2x3 # 20 2x112x22 2x3# 10 z5 2x12 x21 2x3
sx1,x2, . . . , xn, s1, s2, . . . , smd 5 s0, 0, . . . , 0, b1, b2, . . . , bmd.
The Simplex Method (Standard Form)
To solve a linear programming problem in standard form, use the following steps.
1. Convert each inequality in the set of constraints to an equation by adding slack variables.
2. Create the initial simplex tableau.
3. Locate the most negative entry in the bottom row. The column for this entry is called the entering column. (If ties occur, any of the tied entries can be used to determine the entering column.)
4. Form the ratios of the entries in the “b-column” with their corresponding positive entries in the entering column. The departing row corresponds to the smallest non- negative ratio (If all entries in the entering column are 0 or negative, then there is no maximum solution. For ties, choose either entry.) The entry in the departing row and the entering column is called the pivot.
5. Use elementary row operations so that the pivot is 1, and all other entries in the entering column are 0. This process is called pivoting.
6. If all entries in the bottom row are zero or positive, this is the final tableau. If not, go back to Step 3.
7. If you obtain a final tableau, then the linear programming problem has a maximum solution, which is given by the entry in the lower-right corner of the tableau.
biyaij .
Basic
x1 x2 x3 s1 s2 s3 b Variables
2 1 0 1 0 0 10 s1
1 2 0 1 0 20 s2
0 1 2 0 0 1 5 s3 ← Departing
1 0 0 0 0
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
2 1 0 1 0 0 10 s1 ← Departing
1 3 0 0 1 1 25 s2
0 1 0 0 x3
2 0 0 0 1 5
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
1 0 0 0 5 x1
0 0 1 1 20 s2
0 1 0 0 x3
0 3 0 1 0 1 15
This implies that the optimal solution is
and the maximum value of z is 15.
Occasionally, the constraints in a linear programming problem will include an equation.
In such cases, we still add a “slack variable” called an artificial variable to form the ini- tial simplex tableau. Technically, this new variable is not a slack variable (because there is no slack to be taken). Once you have determined an optimal solution in such a problem, you should check to see that any equations given in the original constraints are satisfied.
Example 3 illustrates such a case.
E X A M P L E 3 The Simplex Method with Three Decision Variables
Use the simplex method to find the maximum value of
Objective function
z5 3x11 2x21 x3
sx1, x2, x3, s1, s2, s3d 5 s5, 0, 52, 0, 20, 0d
5 2 1 2 1
2
212 5
2
1 2 1
2
22
5 2 1 2 1
2
22 22
22
subject to the constraints
where and
Solution Using the basic feasible solution
the initial simplex tableau for this problem is as follows. (Note that is an artificial vari- able, rather than a slack variable.)
Basic
x1 x2 x3 s1 s2 s3 b Variables
4 1 1 1 0 0 30 s1 ← Departing
2 3 1 0 1 0 60 s2
1 2 3 0 0 1 40 s3
0 0 0 0
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
1 0 0 x1
0 1 0 45 s2 ← Departing
0 0 1 s3
0 0 0
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
1 0 0 3 x1
0 1 0 18 x2
0 0 1 1 s3
0 0 0 0 45
This implies that the optimal solution is
and the maximum value of z is 45. (This solution satisfies the equation given in the con- straints because 4s3d 1 1s18d 1 1s0d 5 30.d
sx1, x2, x3, s1, s2, s3d 5 s3, 18, 0, 0, 0, 1d
1 2 1 2
2107 1 10 12
5
2
215 5 1 5
2101 3 10 1 5
45 2 3
214 4
254
65
214 2 11
4 7 4
212 1 2 5 2
15 2 1
4 1 4 1 4
21 22 23
s1 sx1, x2, x3, s1, s2, s3d 5 s0, 0, 0, 30, 60, 40d
x3 $ 0.
x1 $ 0, x2$ 0, 2x11 2x21 3x3# 40 2x11 3x213x3# 60 4x113x213x35 30
Applications
E X A M P L E 4 A Business Application: Maximum Profit
A manufacturer produces three types of plastic fixtures. The time required for molding, trimming, and packaging is given in Table 9.1. (Times are given in hours per dozen fixtures.)
TABLE 9.1
Process Type A Type B Type C Total time available
Molding 1 2 12,000
Trimming 1 4,600
Packaging 2,400
Profit $11 $16 $15 —
How many dozen of each type of fixture should be produced to obtain a maximum profit?
Solution Letting and represent the number of dozen units of Types A, B, and C, respec- tively, the objective function is given by
Profit
Moreover, using the information in the table, we construct the following constraints.
(We also assume that and ) Now, applying the simplex method with the basic feasible solution
we obtain the following tableaus.
Basic
x1 x2 x3 s1 s2 s3 b Variables
1 2 1 0 0 12,000 s1 ← Departing
1 0 1 0 4,600 s2
0 0 1 2,400 s3
0 0 0 0
↑ Entering
215 216 211
1 2 1 3 1 2
2 3 2 3
3 2
sx1, x2, x3, s1, s2, s3d 5 s0, 0, 0, 12,000, 4,600, 2,400d x3$ 0.
x1$ 0, x2$ 0,
1
2x1113x2112x3# 12,400
2
3x1123x2132x3# 14,600
2
3x11 2x2132x3# 12,000
5 P 5 11x11 16x21 15x3. x3
x1, x2,
1 2 1
3 1
2
2 3 2
3
3 2
Basic
x1 x2 x3 s1 s2 s3 b Variables
1 0 0 6,000 x2
0 1 0 600 s2
0 0 1 400 s3 ← Departing
0 −3 8 0 0 96,000
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
0 1 0 5,400 x2
0 0 1 200 s2 ← Departing
1 0 0 3 1,200 x1
0 0 0 9 99,600
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 b Variables
0 1 0 1 0 5,100 x2
0 0 1 4 800 x3
1 0 0 0 6 600 x1
0 0 0 6 3 6 100,200
From this final simplex tableau, we see that the maximum profit is $100,200, and this is obtained by the following production levels.
Type A: 600 dozen units Type B: 5,100 dozen units Type C: 800 dozen units
R E M A R K: In Example 4, note that the second simplex tableau contains a “tie” for the minimum entry in the bottom row. (Both the first and third entries in the bottom row are ) Although we chose the first column to represent the departing variable, we could have chosen the third column. Try reworking the problem with this choice to see that you obtain the same solution.
E X A M P L E 5 A Business Application: Media Selection
The advertising alternatives for a company include television, radio, and newspaper advertisements. The costs and estimates for audience coverage are given in Table 9.2 23.
23 24 223
232 13
234 2
212 3 4
21 216
1 4
232 3
4 3 8
23
216 1 4 1
3
213 1 2 1
3
1 2 3 4 1
2
TABLE 9.2
Television Newspaper Radio Cost per advertisement $ 2,000 $ 600 $ 300 Audience per advertisement 100,000 40,000 18,000
The local newspaper limits the number of weekly advertisements from a single company to ten. Moreover, in order to balance the advertising among the three types of media, no more than half of the total number of advertisements should occur on the radio, and at least 10%
should occur on television. The weekly advertising budget is $18,200. How many adver- tisements should be run in each of the three types of media to maximize the total audience?
Solution To begin, we let and represent the number of advertisements in television, news- paper, and radio, respectively. The objective function (to be maximized) is therefore
Objective function
where and The constraints for this problem are as follows.
A more manageable form of this system of constraints is as follows.
Thus, the initial simplex tableau is as follows.
Basic
x1 x2 x3 s1 s2 s3 s4 b Variables
20 6 3 1 0 0 0 182 s1 ← Departing
0 1 0 0 1 0 0 10 s2
1 0 0 1 0 0 s3
1 1 0 0 0 1 0 s4
0 0 0 0 0
↑ Entering
218,000 240,000
2100,000 29
21 21
29x116x213x3# 180 2x126x213x3# 180 20x11 6x21 3x3# 110 20x11 6x21 3x3# 182
2000x11 600x21 300x3$ 0.1sx11 x21 x3d 2000x11 600x21 300x3# 0.5sx11 x21 x3d 2000x11 600x21 300x3# 10 2000x11 600x21 300x3# 18,200
x3$ 0.
x1$ 0, x2$ 0,
z5 100,000x11 40,000x21 18,000x3
x3 x1, x2,
}
ConstraintsNow, to this initial tableau, we apply the simplex method as follows.
Basic
x1 x2 x3 s1 s2 s3 s4 b Variables
1 0 0 0 x1
0 1 0 0 1 0 0 10 s2← Departing
0 0 1 0 s3
0 0 0 1 s4
0 5,000 0 0 0 910,000
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 s4 b Variables
1 0 0 0 x1
0 1 0 0 1 0 0 10 x2
0 0 1 0 s3← Departing
0 0 0 1 s4
0 0 5,000 10,000 0 0 1,010,000
↑ Entering
Basic
x1 x2 x3 s1 s2 s3 s4 b Variables
1 0 0 0 4 x1
0 1 0 0 1 0 0 10 x2
0 0 1 0 14 x3
0 0 0 1 12 s4
0 0 0 0 1,052,000
From this tableau, we see that the maximum weekly audience for an advertising budget of
$18,200 is
Maximum weekly audience
and this occurs when and We sum up the results here.
Number of
Media Advertisements Cost Audience
Television 4 $ 8,000 400,000
Newspaper 10 $ 6,000 400,000
Radio 14 $ 4,200 252,000
Total 28 $18,200 1,052,000
x35 14.
x15 4, x25 10, z5 1,052,000
60,000 23 272,000
23 118,000
23
24723
211823 8
23
20 23 14 23 1
23
2233
2239 1
23
23,000
449
23710 10 9
20 47
20
161 10 7
10 1
20 23
20
61
2103 10 1
20 3
20
23,000 210,000
819 10 9
20 47 20 37
10
91 10 1
20 23
2107 20
91 10 1
20 3
20 3
10
In Exercises 1– 4, write the simplex tableau for the given linear pro- gramming problem. You do not need to solve the problem. (In each case the objective function is to be maximized.)
1. Objective function: 2. Objective function:
Constraints: Constraints:
3. Objective function: 4. Objective function:
Constraints: Constraints:
In Exercises 5–8, explain why the linear programming problem is not in standard form as given.
5. (Minimize) 6. (Maximize)
Objective function: Objective function:
Constraints: Constraints:
7. (Maximize) 8. (Maximize)
Objective function: Objective function:
Constraints: Constraints:
In Exercises 9–20, use the simplex method to solve the given linear programming problem. (In each case the objective function is to be maximized.)
9. Objective function: 10. Objective function:
Constraints: Constraints:
x1, x2$ 10 x1, x2$ 10
3x11 2x2# 12 x114x2# 12
3x11 2x2 # 16 x11 4x2# 18
z5 x11 x2
z5 x11 2x2
x1, x2, x3 $ 0
x1, x2 $ 0 2x11x213x3# 0
2x11 x2 $ 6 2x11 x22 2x3 $ 1
x11 x2 $4 2x11 x21 3x3 # 5
z5 x11 x2
z5 x11 x2
x1, x2 $ 20 2x122x2 # 21 x1, x2 $ 0
2x11 2x2 # 26 x11 2x2 # 4
z5 x11 x2
z5 x11 x2
x1, x2 $ 20 x1, x2, x3 $ 10
2x113x2 # 20 x11 2x21 x3 # 18
2x12 3x2 # 26 x11 2x21 x3# 12
z5 6x12 9x2
z5 2x11 3x21 4x3
x1, x2$ 0 2x1, x2 $ 0
x12 x2 # 1 2x11 x2# 5
x11 x2 # 4 2x11 x2 # 8
z5 x11 3x2
z5 x11 2x2
SECTION 9.3 ❑ EXERCISES
11. Objective function: 12. Objective function:
Constraints: Constraints:
13. Objective function: 14. Objective function:
Constraints: Constraints:
15. Objective function: 16. Objective function:
Constraints: Constraints:
17. Objective function: 18. Objective function:
Constraints: Constraints:
19. Objective function:
Constraints:
20. Objective function:
Constraints:
x1, x2, x3, x4$ 70 2x11 3x21 2x31 6x4# 72 2x11 3x21 2x31 5x4# 50 2x113x21 3x31 4x4# 60 z5 x11 2x21 x32 x4
x1, x2, x3, x4$ 40 x113x21 7x31 x4# 42 x11 2x21 3x31 x4# 24 z5 x11 2x22 x4
x1, x2, x3$ 50 x1, x2, x3$ 30
2x11x21 6x3# 54 2x112x21 3x3# 32
2x11 x21 3x3# 75 2x11 2x213x3# 25
2x11 x213x3# 59 2x112x22 3x3# 40
z5 2x11 x21 3x3
z5 x12 x21 x3
x1, x2$ 40 x1, x2, x3, x4$ 0
3x11 4x2# 48 2x11 4x21 5x31 2x4# 8
3x11 2x2# 28 2x11 6x214x31 5x4# 3
3x11 2x2# 60 8x11 3x21 4x315x4# 7
z5 x1
z5 3x11 4x21 x31 7x4
x1, x2$ 10 x1, x2$ 40
2x123x2# 12 3x11 7x2# 42
2x11 3x2# 15 3x117x2# 10
z5 x11 2x2
z5 4x11 5x2
x1, x2, x3$ 40
x1, x2, x3$ 0 6x123x21 3x3# 42
2x11 2x3# 5 2x11 3x223x3# 42
2x11 2x2# 8 2x12 4x213x3# 42
z5 x12 x21 2x3
z5 5x11 2x21 8x3
21. A merchant plans to sell two models of home computers at costs of $250 and $400, respectively. The $250 model yields a profit of $45 and the $400 model yields a profit of $50. The merchant estimates that the total monthly demand will not exceed 250 units. Find the number of units of each model that should be stocked in order to maximize profit. Assume that the merchant does not want to invest more than $70,000 in computer inventory. (See Exercise 21 in Section 9.2.) 22. A fruit grower has 150 acres of land available to raise two
crops, A and B. It takes one day to trim an acre of crop A and two days to trim an acre of crop B, and there are 240 days per year available for trimming. It takes 0.3 day to pick an acre of crop A and 0.1 day to pick an acre of crop B, and there are 30 days per year available for picking. Find the number of acres of each fruit that should be planted to maximize profit, as- suming that the profit is $140 per acre for crop A and $235 per acre for B. (See Exercise 22 in Section 9.2.)
23. A grower has 50 acres of land for which she plans to raise three crops. It costs $200 to produce an acre of carrots and the profit is $60 per acre. It costs $80 to produce an acre of celery and the profit is $20 per acre. Finally, it costs $140 to produce an acre of lettuce and the profit is $30 per acre. Use the simplex method to find the number of acres of each crop she should plant in order to maximize her profit. Assume that her cost cannot exceed $10,000.
24. A fruit juice company makes two special drinks by blending apple and pineapple juices. The first drink uses 30% apple juice and 70% pineapple, while the second drink uses 60%
apple and 40% pineapple. There are 1000 liters of apple and 1500 liters of pineapple juice available. If the profit for the first drink is $0.60 per liter and that for the second drink is
$0.50, use the simplex method to find the number of liters of each drink that should be produced in order to maximize the profit.
25. A manufacturer produces three models of bicycles. The time (in hours) required for assembling, painting, and packaging each model is as follows.
Model A Model B Model C
Assembling 2 2.5 3
Painting 1.5 2 1
Packaging 1 0.75 1.25
The total time available for assembling, painting, and packag- ing is 4006 hours, 2495 hours and 1500 hours, respectively.
The profit per unit for each model is $45 (Model A), $50 (Model B), and $55 (Model C). How many of each type should be produced to obtain a maximum profit?
26. Suppose in Exercise 25 the total time available for assem- bling, painting, and packaging is 4000 hours, 2500 hours, and 1500 hours, respectively, and that the profit per unit is $48 (Model A), $50 (Model B), and $52 (Model C). How many of each type should be produced to obtain a maximum profit?
27. A company has budgeted a maximum of $600,000 for adver- tising a certain product nationally. Each minute of television time costs $60,000 and each one-page newspaper ad costs
$15,000. Each television ad is expected to be viewed by 15 million viewers, and each newspaper ad is expected to be seen by 3 million readers. The company’s market research department advises the company to use at most 90% of the advertising budget on television ads. How should the advertising budget be allocated to maximize the total audi- ence?
28. Rework Exercise 27 assuming that each one-page newspaper ad costs $30,000.
29. An investor has up to $250,000 to invest in three types of in- vestments. Type A pays 8% annually and has a risk factor of 0. Type B pays 10% annually and has a risk factor of 0.06.
Type C pays 14% annually and has a risk factor of 0.10. To have a well-balanced portfolio, the investor imposes the fol- lowing conditions. The average risk factor should be no greater than 0.05. Moreover, at least one-fourth of the total portfolio is to be allocated to Type A investments and at least one-fourth of the portfolio is to be allocated to Type B invest- ments. How much should be allocated to each type of invest- ment to obtain a maximum return?
30. An investor has up to $450,000 to invest in three types of investments. Type A pays 6% annually and has a risk factor of 0. Type B pays 10% annually and has a risk factor of 0.06.
Type C pays 12% annually and has a risk factor of 0.08. To have a well-balanced portfolio, the investor imposes the following conditions. The average risk factor should be no greater than 0.05. Moreover, at least one-half of the total portfolio is to be allocated to Type A investments and at least one-fourth of the portfolio is to be allocated to Type B invest- ments. How much should be allocated to each type of investment to obtain a maximum return?
31. An accounting firm has 900 hours of staff time and 100 hours of reviewing time available each week. The firm charges
$2000 for an audit and $300 for a tax return. Each audit requires 100 hours of staff time and 10 hours of review time, and each tax return requires 12.5 hours of staff time and 2.5 hours of review time. What number of audits and tax returns will bring in a maximum revenue?
